I am facing a physics problem that requires me to solve $$ 0 = -gt - \frac{F}{\alpha}\cdot \log\left( 1 - \frac{\alpha t}{M+m} \right) $$ with $M=5~kg$, $m=10~kg$, $\alpha = 1 kg/s$ and $F = 0.9(M +m)g$. Via computer trial and error I was able to find that $t = 0$ and $t \approx 2.9$ are solutions. However, how can I find the solutions on a mathematical way? Since the parameters are not necessary for the mathematics, I broke it down to this for you:

$$ 0 = ax + b \cdot \log(1 + cx) $$

Note: In the exercise following hint is given: Note that $\alpha t \ll M+m$. It would be helpful already, if you could help me use that hint. Because that hint means (from my point of view) that $\frac{\alpha t}{M+m} \approx 0$. However, it would follow $\log(1 - 0) = 0$ and from that $t = 0$. This is one solutions, but how to find the second (more important) solution?

  • $\begingroup$ You will have to solve this numerically. $\endgroup$ Dec 14 '17 at 11:00
  • $\begingroup$ Anyway, with the hint, it means that the $\log$ term can be neglected so you have a simple linear equation to solve only. $\endgroup$ Dec 14 '17 at 11:01
  • $\begingroup$ @KarnWatcharasupat As I wrote, this linear term $ 0 = -gt$ would yield $t =0$, but how to find $t = 2.9$? $\endgroup$ Dec 14 '17 at 11:03

$$0=ax+b\ln(1+cx) \tag 1$$ Note : If $\quad 0=ax+\beta\log_{10}(1+cx)\quad\to\quad b=\beta\ln(10)$

$\ln(1+cx)=-\frac{a}{b}x\quad\to\quad 1+cx=e^{-\frac{a}{b}x}\quad\to\quad (1+cx)e^{-\frac{a}{bc}}=e^{-\frac{a}{b}x-\frac{a}{bc}}$


$ (\frac{a}{b}x+\frac{a}{bc})e^{\frac{a}{b}x+\frac{a}{bc}}=\frac{a}{bc}e^{\frac{a}{bc}}$

The solution of $\quad Xe^X=Y\quad$ is $\quad X=W(Y)\quad$ Lambert $W$ function. http://mathworld.wolfram.com/LambertW-Function.html

With $X=(\frac{a}{b}x+\frac{a}{bc})$ and $Y=\frac{a}{bc}e^{\frac{a}{bc}} \quad\to\quad\frac{a}{b}x+\frac{a}{bc}=W\left(\frac{a}{bc}e^{\frac{a}{bc}} \right)$

The solution of the equation $(1)$ is : $$x=-\frac{1}{c}+\frac{b}{a}W\left(\frac{a}{bc}e^{\frac{a}{bc}} \right)$$

The solution cannot be expressed with a finite number of elementary functions. Many equations requires some special functions to have the solution expressed on closed form. This is the case of Equation $(1)$ which solving involves the special function $W$.

A paper for general public about special functions : https://fr.scribd.com/doc/14623310/Safari-on-the-country-of-the-Special-Functions-Safari-au-pays-des-fonctions-speciales

If you don't want to use a special function, approximate solutions are commonly computed in using numerical calculus. That is the method recommended in your case insofar the numerical values of the parameters are known.


Using Taylor series for $\ln(1+x)$ plugging $x=-t$ we get $-x-\frac{x^2}2-\frac{x^3}{3}+\cdots$ $$0=-gt-\frac{F}{\alpha}\cdot(-\frac{\alpha t}{M+m}-\frac{\alpha^2t^2}{2(M+m)^2}-\cdots)$$ Now the more terms of the taylor expansion you take the better the approximation, since we can solve a quadratic and we know that $t=0$ is a solution we can take three terms. $$gt=F(\frac{t}{M+m}+\frac{\alpha t^2}{2(M+m)^2}+\frac{\alpha^2t^3}{3(M+m)^3})$$ $$gt=0.9g(M+m)(\frac{t}{M+m}+\frac{\alpha t^2}{2(M+m)^2}+\frac{\alpha^2t^3}{3(M+m)^3})$$ Since $t=0$ zero is a solution for $t\neq 0$ we divide by $gt$ to get $$1=0.9+\frac{0.9t}{2(M+m)}+\frac{0.9t^2}{3(M+m)^2}$$ From this $t\approx 2.94727$ (because the other solution is negative). Since $\frac{\alpha t}{M+m}\approx 0.2$ is not small enough the approximation isn't good enough, in case you can solve a $3$rd degree polynomial then taking $4$ terms gives approximately $t\approx 2.904$, larger degree of polynomial equals better approximation.


From your parameters, we see that $a\neq 0$, $b\neq 0$ and $c\neq 0$.

The solution of your problem is given by using the Lambert $W$ function

$$x = \dfrac{b}{a}W\left[\dfrac{a}{bc}\exp\left(\dfrac{a}{bc}\right) \right]-\dfrac{1}{c}$$


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